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Need Mathmatical help from those who can offer itOkay, so here's my situation. I'm sterling scholar for math at my high school(yes pimp, that means you can go to http://www.cvhs.iron.k12.ut.us and look at my pictures for sterling scholar and cross country). For those who don't know what sterling scholar is it's just an academic competition thing.
This is where you guys come in. Every year our math teacher has the math sterling scholar make a presentation to our calculus class and some kind of demonstration thing with it to put in the portfolios we make for this competition. I intend to make two identical spud guns, one combustion, one pneumatic, and estimate muzzle velocities for them using math(I'm familiar with calculus so most equations won't faze me) and also find the point where the two will have the same muzzle velocity. I'll be making a propane metered combustion w/ chamber fan and a barrel sealing piston valved pnuematic. Any equations you guys could post or places that I could look to find the info I'll need would be helpful. Thanks, Squeaks P.S. Please don't refer me to the GGDT or any like devices. The whole point is to show that I can do it without said devices.
"Nine out of ten Americans agree that out of ten Americans one will always disagree with the other nine."
Collin Mockery Who's Line is it Anyway Borrow money from a pessimist, he won't expect it back.
When you say identical, what exactly does that mean?
Same barrel length/diameter & the same chamber volume for both cannons? Basic predictions can be made with: v^2 = u^2 + 2*a*s In any case, doubling the acceleration, means that muzzle velocity will increase by about 40% (It increases by a factor equal to the square root of 2) To double the acceleration, you need double the force, and thus twice the pressure. With a weighty projectile and good flow, then you can assume the pressure in the barrel is constant, and is roughly defined by the average of the pressures at the base of the barrel and the muzzle: At the base of the barrel, the pressure is roughly equal to the Initial pressure in your chamber (after ignition in a combustion, immediately after valve opening in a pneumatic) You can quite easily work out what the pressure will be at the muzzle for a pnuematic. It's about Initial pressure * chamber volume / (chamber volume+barrel volume) I'm not sure about what the combustion maestros will say about combustion pressures along the barrel. Knowing the average pressure allows the average acceleration to be found, and then from that, a good estimate of muzzle velocity. If you're just skipping over the basic predictions, then the maths is mindnumbingly complex  and I don't have the time to expalin it now.
Novacastrian: How about use whatever the heck you can get your hands on?
frankrede: Well then I guess it won't matter when you decide to drink bleach because your out of koolaid. ...I'm sorry, but that made my year.
Sorry I can't contribute to the matter. I'm in precalculus and honest to God I don't know what the hell is going on. My school doesn't award the math credits I need if you retake classes, so in essence I'm stuck. By the way, you know that hot Laci chick that used to go to school up there? [Not sure if thats how you spell it.] Yeah, shes in my math class.
[Score!] Its like you read my mind! Although, I'm afraid you will not see me in list of sterling scholars anytime soon. Maybe for football though. The headline will probably read something along the lines of 'CVHS loses again, when in opposite of heaven are they going to get a new coach!'. I don't know if your intent is to solely model the discussed launchers, just giving you a reason to build them. Anyway, I got a 8x12 metal lathe I could help you out with, also a shooting chronograph to help you ruff those slightly incorrect numbers into place. :p From what I understand projectile friction is a bit hard to correctly asses, so thats one of the things your going to have and tweak to produce sufficient results. As always, the best of luck!
Okay, thanks joan, you were correct with the assumption of same chamber/barrel volume and same chamber/barrel legnth and diameter. That equation looks like it will be just what I need, just a little clarification quickly.
Could you kindly explain some of your variables. I get that it's velocity squared equals something squared plus two acceleration times something. Also, to find the accleration I'm going to have to use A=F/M. So I'll take the pressure in pascals and multiply it by the surface area then divide by the mass of my projectile right? Lastly, where am I going to subjtract the resistance from? Thanks for the offer with the lathe pimp, but I'm going to keep this one small and simple so I won't need it. As to the chrony though I might need it. After making my predicitions I'm going to check it using the GGDT and get actual chrony data. I'm going to try and get one from my local paintball shop for the day but I'm not sure if they'll let me. Last thing, anyone know how to find the pressure after combustion of propane?
"Nine out of ten Americans agree that out of ten Americans one will always disagree with the other nine."
Collin Mockery Who's Line is it Anyway Borrow money from a pessimist, he won't expect it back.
For v^2 = u^2 +2*a*s
V is final velocity u is initial velocity (as your spud is initially stationary, this can be ignored) a is acceleration s is displacement (your barrel length). Acceleration is found via: a = f/m (as you said). To find the force, you'd need the pressure in pascals (averaged as explained before), times surface area in square metres. You subtract the resistance (in newtons) from this figure, then feed it into a = f/m. The model isn't 100% accurate because it doesn't account for gas flow, but it's estimates aren't far off for heavy projectiles and good flow. Spuds and piston valves are within a fairly accurate zone. I understand that propane under perfect conditions combusts at 102 psig (About 70 kPa), but in a spudcannon it's rather less. I don't know exactly what it is though, and different people will tell you everything from 40 psig to up to 7080 psig. BoilingLeadBath is probably the best person to ask, given all of his work on the EVBEC spreadsheets.
Novacastrian: How about use whatever the heck you can get your hands on?
frankrede: Well then I guess it won't matter when you decide to drink bleach because your out of koolaid. ...I'm sorry, but that made my year.
The problem with estimating the performance of a combustion gun from a physics standpoint is, we dont know very much about the dynamics of combustion in a changing volume such as a combustion chamber. At the moment, the best we can do is use data from Latke's chronometer tests to estimate muzzle velocity and energy by means of comparison. This is quite accurate, but only if you have a launcher that is similar in size and configuration to the Latke test launchers.
Jimmy from the Spudtech forums is working on a physics based model of a combustion gun, which when completed should yield much more accurate results than what we have now.
People should not be afraid of their governments. Governments should be afraid of their people.
The muzle velocity of the 3/4" latke combustion launcher can be predicted with incredible accuracy by the following formula:
=SQRT(719550/(X+1)*X119450*X)+5.865*X Where X is the B:C ratio and 719550, 119450, and 5.865 are constants found by fitting the curve. Now, you'll definitly want to explain the equation. You see, what we have here is the absolute pressure [constant/(x+1)] minus the atmospheric pressure [constant] integrated over barrel length [the extra X terms] In other words, this is the force times distance, or energy. We then take the square root of that, to turn energy into velocity. (show how you get that procedure by solving the classic physics equation, E=1/2MV^2, for V) I'm not sure exactly what the 5.865X represents, but I'm putting my on it acctualy being a X^2 term inside the square root and on it representing the heat lost to the barrel/chamber walls. Given that, you can apply the equations of the earlier EVBECs  as recent as V1.3, acctualy  and that should be close enough for your clasmates.
Did you use some sort of regression to find that equation? Just wondering.
Here's a link to a chrony design that you should find easy to build. http://home.earthlink.net/~jimsluka/Jims_chrono.html It's basically two photodiodes mounted at known intervals in front of the barrel. The cleverness in it is that the output of the chrony is to the speaker input on a computer. Download the right freeware and instant oscilloscope. The projectile passing the photodiodes interrupts the light and gives you two sharp dips in the DC signal. The math is trivial. You could probably record a lot of data to tape or digital recorder and analyze at your leisure. I know it's a math competition but maybe there are some bonus points for good applied kitchen sink science? Good luck!
The form of the formula was predicted through a bit of applied physics (see description of pressure integrated over barrel length), and the constants found via trialanderror minimalization of the chi square (or whatever it's called).
There are curve fitting programs that will let you fit a polynomial to any set of experimental data with arbitrary precision. If the formula can be expressed as a polynomial of arbitrary order, tell the program to use that order in the curvefitting. Typically it will spit out both the constants and the "goodness" of the fit based on the sum of squares. So:
Good luck on the project.
Why the hell should I define it as a polynomial function (I'm assuming you meant of the X^Y type) when a much simpler  and physical meaningful  formula exists?
...anyway, I believe I did fit the datapoints to a polynomial at one point in time. I believe I had to make it like 4th or 5th order to get reasonably close. (and we don't want to simply find the minimal sumofsquares; different datapoints have different standard deviations, so we have to weight them accordingly.) Anyway, I figure a picture of what I mean by "extremely well" is in order: <a href="http://img.photobucket.com/albums/v611/car2/Modeling.png">graph</a> Don't pay attention to the purple line labeled "3/4 attempt 2"... it's only in there to demonstrate what happens when you remove the 5.865X term.
Wow! That's a dang good fitting graph. Is that measurement in feet per second or miles per hour? Oh yeah, one last thing, is that assuming a 4.2% mix of propane?
"Nine out of ten Americans agree that out of ten Americans one will always disagree with the other nine."
Collin Mockery Who's Line is it Anyway Borrow money from a pessimist, he won't expect it back.
Output is in feet per second.
<a href="http://www.burntlatke.com/jpg600/cbtestdata.gif">Test conditions</a> They don't state their %mix, but using ratcalc with their given chamber volume, meter pressure, and meter volume, it seems that it's roughly 4.2% propane.
<flame> I said pick a polynomial of order that matches the model. If you have to go 4th or 5th order to get a good fit to the data, and you're confident of your model, then your data has too much scatter. And there are curve fitting approaches that use a weighted sum of squares. If this were a peerreviewed academic journal, you'd be back at the lab bench improving your experimental technique, not swearing in public at the reviewer. But hey, it's a BBS for a hobby with a big Bubba Factor so lighten up. </flame>
 
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